Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality: $$2^{2^{2^{2^{\mathfrak{n}}}}} \lt \left(2^{2^{2^{2^{\mathfrak{n}}}}}\right)^2 = 4^{2^{2^{2^{\mathfrak{n}}}}}$$ Where "finite" is meant in the strictest Tarskian sense: in bijection with a finite ordinal.

My question is a bit vague since "simple" has a lot of interpretations but here it is:

Is there a simpler cardinal inequality that is equivalent to finiteness and uses only cardinal exponentiation?